just wondering if someoen could help me with this limit..:

lim arctan[(x^2 - 4)/(3x^2-6x)]
x->2

My first step in doing limits is to sub in the approach value, this will give me Lim arctan (0/0)

This just about guarantees that the algebraic expression will factor and the offending factor will cancel.
Sure enough
lim arctan[(x^2 - 4)/(3x^2-6x)] x->2
= lim arctan[(x+2)(x-2)]/[3x(x-2)]
=lim arctan[(x+2)]/[3x]
= lim arctan(2/3)

now use your calculator, set to radian mode, to find
lim arctan[(x^2 - 4)/(3x^2-6x)] x->2
= .588

thanks for the reply... the question asks me to evaluate the limit using continuity, how do i justify my answer with continuity?

lim x^2+3x-4=6

x->2

Sure, I can help you with that limit. To find the limit of the given expression, we can use some algebraic manipulation and trigonometric identities.

Let's start by simplifying the expression inside the arctan function:

arctan[(x^2 - 4)/(3x^2 - 6x)]

Factor out an x^2 from both the numerator and denominator:

arctan[(x^2(1 - 4/x^2))/(x^2(3 - 6/x))]

Cancel out the common factors of x^2:

arctan[(1 - 4/x^2)/(3 - 6/x)]

Now, we can take the limit as x approaches 2:

lim(x->2) arctan[(1 - 4/x^2)/(3 - 6/x)]

Next, let's simplify the expression inside the arctan function:

lim(x->2) arctan[(1 - 4/4)/(3 - 6/2)]

Simplifying further:

lim(x->2) arctan[(1 - 1)/(3 - 3)]

Now, we have:

lim(x->2) arctan[0/0]

To proceed further, we need to apply L'Hôpital's rule, which states that if we have a limit of the form 0/0 or ∞/∞, we can differentiate the numerator and denominator until we get an indeterminate form that is easier to evaluate.

Differentiating the numerator and denominator:

lim(x->2) [d/dx (1 - 1)/(d/dx (3 - 3))]

Simplifying further:

lim(x->2) [0/0]

Applying L'Hôpital's rule again:

lim(x->2) [d/dx (0/0)]

Differentiating once more:

lim(x->2) [d^2/dx^2 (0/0)]

We can continue this process until we obtain a determinate form or conclude that the limit does not exist.

Following the same process, we can differentiate again:

lim(x->2) [d^3/dx^3 (0/0)]

Once again,

lim(x->2) [d^4/dx^4 (0/0)]

And so on...

After several differentiations, we still end up with the indeterminate form of 0/0. This suggests that L'Hôpital's rule is not applicable, and we need to try a different approach.

To evaluate the final limit, we can make use of the fact that the arctan function has a specific limit as its argument approaches zero:

lim(y->0) arctan(y) = 0

By substituting y = (1 - 4/x^2)/(3 - 6/x), we can rewrite the limit as:

lim(x->2) arctan[(1 - 4/x^2)/(3 - 6/x)] = lim(y->0) arctan(y)

Now, we can see that the limit becomes:

lim(y->0) arctan(y) = arctan(0) = 0

Therefore, the limit of the given expression as x approaches 2 is 0.

It's important to note that the steps followed to solve this particular limit may vary depending on the specific problem. Always be mindful of the properties and rules associated with limits, as well as any specific algebraic manipulations or trigonometric identities that might be helpful.